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FLUID MECHANICS PLEASE HELP?
Water stands at a depth H in a large open tank whose side walls are vertical. A hole is made in one of the walls at a depth h below the water surface.
At what distance R from the foot of the wall does the emerging stream strike the floor? 2sqrt(h(H-h))
How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole?????
3 Answers
- Luigi 74Lv 71 decade agoFavorite Answer
At the hole the speed of the water is : v = sqr{2 g(H-h)} horizontal.
R = v*t = v*sqr{2h/g} = sqr{2g(H-h)*2h/g} = 2 sqr{h(H-h)}
h' = h
- Anonymous1 decade ago
for the question just use ur formulas!
for the second question: depth H
- Anonymous1 decade ago
go to the internet and type it in to find out the answer
Source(s): internet